Introduction to Continuity Of Functions - At A Glance:

We restricte the values of x until we got what we wanted, making sure that

x got the same distance away from c to either side, and that

we didn't restrict x to just equal c, since then we would find a dot.

Enter the Greek letters, frat boys rejoice.

In symbols,

We are given a continuous function f and a value c.

We pick a real number ε>0 (epsilon) that said how far we wanted to let f(x) move away from f(c).

We restrict the values of x until we got what we wanted, ending with c - δ

The continuous function guarantee says that no matter what ε > 0 we pick, we'll be able to find a δ > 0.

We use the Greek letters to specify how close various things are to one another. We use the letter ε to specify how close we want f(x) and f(c) to be. The definition of continuity says for any ε we can find an appropriate δ such that if x is within δ of c, we find our desired value.

We have a great recipe for cooking up δ with the ingredients f, c, and ε. First we combine flour, baking soda, and salt in a bowl, then we...no wait, sorry, that's the recipe for Nestle Tollhouse Cookies.

Write down the inequalityf(c) - ε < f(x) < f(c) + εand fill in whatever we are given for c, f, and ε.

Solve the inequality for x.

Subtract c from all parts of the inequality and find δ.

The super-formal definition of continuity says:

The function f is continuous at c if for any real ε > 0 there exists a real δ > 0 such that if |x - c| < δ, then|f(x) - f(c)| < ε.

To translate, if f is continuous at c we can pick any real ε > 0 and say we want to have f(x) and f(c) within ε of each other. In symbols, we write this|f(x) - f(c)| < ε.

This is the same thing as saying-ε < f(x) - f(c) < εwhich is the same thing as sayingf(c) - ε < f(x) < f(c) + ε.

In pictures,

Since f is continuous, we have a guarantee that we can find some real δ > 0 such that if x is within δ of c, we find what we want.In symbols, we say|x - c| < δwhich means the same thing as-δ, which means the same thing asc - δ.

Example 2

Let f(x) = x2, c = 2, and ε = 0.1. Find an appropriate value of δ>0 as guaranteed by the continuity of f at 2.

The first thing we want to do is set up an inequality that looks likef(c) - ε < f(x) < f(c) + ε.In this case, we have c = 2 and sof(c) = 22 = 4. Since ε = 0.1, we find 4 - 0.1<f(x)< 4 + 0.1, or 3.9 < x2 < 4.1.

Now we need to solve the inequality.Taking square roots of all parts of the inequality, we find Evaluating the square roots and rounding a lot,we need to have 1.97 < x < 2.02.

To find δ, subtract c = 2 from all parts of the inequality. 0.03 < x - 2 < 0.02. This says x can move to 0.03 away from 2 to the left, but only 0.02 away from 2 to the right. Since we want to restrict the values of x so c = 2 is in the middle, we choose the smaller bound, and go with δ = 0.02. We can check this answer. As long as x is within δ of c, that is, 2 - 0.02 < x < 2 + 0.02, we find (1.98) 22<(2.02) 2. Then 3.9 < 3.9204 < x2 < 4.0804 < 4.1, which means 3.9 < f(x) < 4.1. That's just what we wanted.

Solve the inequality for x.Subtract 1 from each part of the inequality to find 19.9 < 4x < 20.1, then divide each part by 4 to find 4.975 < x < 5.025.

Subtract c from all parts of the inequality and find δ.Since c = 5, we subtract 5 from each part of the inequality to find -0.025 < x-5 < 0.025 therefore |x - 5| < 0.025. Then δ = 0.025 is what we want.